The equation $ f'(x)=f(x)$ admits a solution

Question

let $f :[0,1]→\mathbb R$ be a fixed continous function such that f is differentiable on (0,1) and $ f(0)=f(1)=0$ .then the equation $ f'(x)=f(x)$ admits

1. No solution $x\in (0,1)$
2. More than one solution $x\in (0,1)$
3. Exactly one solution $x\in (0,1)$
4. At least one solution $x\in (0,1)$

Actually, mean value theorem doesn't work for $f'(x)$. The function $f$ has a fixed point and at that point $f'(x)=1$ and at some point $x_0 , f'(x)=0$ (by mean value theorem).but it doesn't get me anywhere! What am I missing?

Obviously 1. Is false since f(x)=0 (also 3.)

2. Is also false since $f(x)=\sin\pi x$

So 4. Is true, but I stuck to prove. Please help.

Answer 1

hint

$$g(x)=f(x)e^{-x}$$

$$g(0)=g(1)$$

Rolle?

Answer 2

Look at $ g(x)=f(x) \cdot \exp (-x) $. Then $ g $ is also differentiable on $ (0,1) $ with derivative $ g'(x) = (f(x)-f'(x)) \exp (-x) $. Noting that $ g(0) = g(1) = 0 $, Rolle's theorem applies and you get a $ c \in (0,1) $ such that $ g'(c) = 0 $. This implies $ f(c) = f'(c) $.